Daily math problems for October 19, 2026 — Descriptive Statistics & Sampling, Euclidean Geometry, Calculus & more

$P(\bar{A})=1-P(A)=1-0.3=0.7$.

$P=2(5+3)=16$ cm.

$f^{\mathrm{\prime }}(x)=2x$ and $f^{\mathrm{\prime }\mathrm{\prime }}(x)=2>0$ for all $x\in \mathbb{R}$.

Since $f^{\mathrm{\prime }\mathrm{\prime }}(x)>0$ everywhere, $f$ is convex (concave up) on all of $\mathbb{R}$ — no inflection point.

$x^2-5x+6=(x-2)(x-3)$.

So ${\displaystyle \frac{x^2-5x+6}{x-2}}=x-3$, with no remainder.

Apply the product rule with $u=(2x-1{)}^4$ and $v=x+1$.

Differentiate each part (chain rule on $u$): $$u^{\mathrm{\prime }}=4(2x-1{)}^3\cdot 2=8(2x-1{)}^3,v^{\mathrm{\prime }}=1.$$

So $$f^{\mathrm{\prime }}(x)=8(2x-1{)}^3(x+1)+(2x-1{)}^4\mathrm{.}$$

Factor out $(2x-1{)}^3$: $$f^{\mathrm{\prime }}(x)=(2x-1{)}^3[8(x+1)+(2x-1)]=(2x-1{)}^3(10x+7)\mathrm{.}$$

Substitute $x=1$, where $2x-1=1$: $$f^{\mathrm{\prime }}(1)=(1{)}^3(10+7)=17.$$

$x^2-4=0\Rightarrow x=\pm 2$. Both vertical asymptotes.

The second equation is $2\times$ the first, so it provides no new information. The system has infinitely many solutions.

$2\sin x\cos x-\sqrt{3}\cos x=0\Rightarrow \cos x(2\sin x-\sqrt{3})=0$.

$\bullet$ $\cos x=0\Rightarrow x\in \text{\hspace{0.17em}⁣}\{{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{3\pi }{2}}\}$

$\bullet$ $\sin x={\displaystyle \frac{\sqrt{3}}{2}}\Rightarrow x\in \text{\hspace{0.17em}⁣}\{{\displaystyle \frac{\pi }{3}},{\displaystyle \frac{2\pi }{3}}\}$

Total: $4$ solutions on $[0,2\pi )$.

$\bullet$ Rewrite in power form using $\ln \sqrt[3]{x^{10}}=\frac{10}{3}\ln x$:

$$f(x)=x^{-7\mathrm{/}3}+\frac{10}{3}\ln x$$

$\bullet$ Differentiate once:

$$f^{\mathrm{\prime }}(x)=-\frac{7}{3}{x}^{-10\mathrm{/}3}+\frac{10}{3x}$$

$\bullet$ Differentiate again:

$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=\frac{70}{9}{x}^{-13\mathrm{/}3}-\frac{10}{3x^2}$$

$\bullet$ Plug in $x=2$ in $f^{\mathrm{\prime }}$ and $x=3$ in $f^{\mathrm{\prime }\mathrm{\prime }}$:

$$f^{\mathrm{\prime }}(2)=-\frac{7}{3\cdot 2^{10\mathrm{/}3}}+\frac{5}{3}$$

$$f^{\mathrm{\prime }\mathrm{\prime }}(3)=\frac{70}{9\cdot 3^{13\mathrm{/}3}}-\frac{10}{27}$$

R2 − R1: $y+2z=8$. R3 − R1: $3y+8z=30$. From the first: $y=8-2z$. Substituting: $3(8-2z)+8z=30\Rightarrow 24+2z=30\Rightarrow z=3$. Then $y=2$, $x=1$. Solution: $(1,2,3)$.